Oriented pipe, midspan stress — SWBPIPE verification

Part VI · VerificationVI–C8

Part VI

Verification

C. Stress recovery · Oriented pipe, midspan stress

A 4.0 m cantilever pipe runs from (0.0, 0.0, 0.0) to (0.0, 4.0, 0.0), along global Y rather than global X. A reference vector (1.0, 0.0, 0.0) fixes its local y axis. It carries a uniform local load of −2.0 N/m and a point force of −4.0 N at midspan. Given its free-end movement in global terms, does the solver recover the same midspan moment and bending stress as for the same pipe laid along X?

Fig. VI–C8.—Oriented pipe, midspan stress

1.Inputs.

Illustrative values, taken from no standard and chosen so the arithmetic can be followed by hand. Most do not describe a real pipe; read them in any consistent set of units.

Quantity	Symbol	Value
Length	L	4.0 m
Elastic modulus	E	1000.0 Pa
Shear modulus	G	400.0 Pa
Area	A	3.0 m²
Second moment about local y	I_y	1.5 m⁴
Second moment about local z	I_z	2.0 m⁴
Torsion constant	J	1.0 m⁴
Uniform local y load	q	−2.0 N/m
Point local y force	P	−4.0 N
Point force station	a/L	0.5
Node 1 global displacement	U_X	−0.04533333 m
Node 1 global rotation	R_Z	0.01466667 rad
Section modulus about y	Z_y	2.5 m³
Section modulus about z	Z_z	2.0 m³
Torsion radius	r	0.5 m

2.Method.

With the given reference vector, local x points along global +Y, local y along global +X and local z along global −Z. The solver transforms the global displacements into local element coordinates before recovering forces. That gives the same local deflection and rotation as the aligned pipe.

From there the station walk is unchanged. The fixed-end shear is V_y,i = 12.0 and moment M_z,i = 24.0; the point force at the station is counted in. The bending stress is the moment over the section modulus.

v_local,y = −0.04533333333333334 m, θ_local,z = −0.014666666666666668 rad(1)

V_y(x) = 12.0 + (−2.0 × 2.0) + (−4.0) = 4.0 N(2)

M_z(x) = 24.0 − 12.0 × 2.0 − (−2.0 × 2.0²/2) − (−4.0 × 0.0) = 4.0 N·m(3)

σ_b,z = M_z / Z_z = 4.0 / 2.0 = 2.0 Pa(4)

σ_axial = N / A = 0.0 / 3.0 = 0.0 Pa(5)

3.Results.

Expected midspan values
Quantity	Expected
Midspan bending moment, M_z	4.0 N·m
Bending stress about z	2.0 Pa
Axial normal stress	0.0 Pa

The tests check that:

The midspan stress recovery completes without being blocked and is labelled as this case’s midspan station.
The midspan bending moment matches 4.0 N·m.
The bending stress about z matches 2.0 Pa.
The axial normal stress is exactly zero.
The result carries the review flags the program attaches to every result.

What it shows. Compared with an independent hand calculation.

Path exercised. The benchmark calls the solver’s components directly: elements, loads, frame solver and stress recovery. It does not go through the program’s own model-to-solve path.

Agreement. The moment and stresses must match the hand calculation to within an absolute difference of 1.0 × 10⁻⁹ in their own units. The zero axial stress, station label and review flag are checked exactly. Long values are shown here to seven significant figures; the tests compare the full values in the record.

For the student

The global rotation about Z is positive, yet the local rotation about z is negative. Nothing has changed physically: the local z axis points along global −Z, so the same rotation reads with the opposite sign. Recovering forces in the wrong frame is a common source of sign errors.

4.Run it yourself.

cd projects/chirality-piping
cargo test --manifest-path validation/benchmarks/stress/Cargo.toml oriented_load_to_stress_fixture_recovers_midspan_bending_from_global_model

Hand calculation: validation/hand_calcs/stress/tp_phys_005_oriented_load_to_stress.md. Test record, with the recorded run of 2026-07-10: stress-tp-phys-005-oriented-load-to-stress.md.